power logsine integral $\int_0^\frac{\pi}{2}\ln^n{(\sin{x})}dx$ It is not hard to find that $\int_0^\frac{\pi}{2}\ln{(\sin{x})}dx=-\frac{\pi}{2}\ln2$.
https://socratic.org/questions/how-do-you-prove-that-the-integral-of-ln-sin-x-on-the-interval-0-pi-2-is-converg
Moreover wolfram can compute that 
 $$\int_0^\frac{\pi}{2}\ln^2{(\sin{x})}dx=\frac{\pi^3+3\pi\ln^24}{24}$$
 $$\int_0^\frac{\pi}{2}\ln^3{(\sin{x})}dx=-\frac{\pi}{16}(12\zeta(3)+\ln^34+\pi^2\ln4)$$
and  $$\int_0^\frac{\pi}{2}\ln^4{(\sin{x})}dx=\pi\zeta(3)\ln{8}+\frac{19\pi^5}{480}+\frac{\pi}{2}\ln^42+\frac{\pi^3}{4}\ln^22$$
How to prove these formulas?
In general, How to evaluate the integral $\displaystyle\int_0^\frac{\pi}{2}\ln^n{(\sin{x})}dx$ ?
Thanks in advance.
 A: $\begin{align}J_n=\int_0^{\frac{\pi}{2}}\ln^n(\sin x)\,dx\end{align}$
Perform the change of variable $y=\sin x$,
$\begin{align}J_n=\int_0^1 \frac{\ln^n(x)}{\sqrt{1-x^2}}\,dx\end{align}$
Perform the change of variable $y=x^2$,
$\begin{align}J_n=\frac{1}{2^{n+1}}\int_0^1 \frac{x^{-\frac{1}{2}}\ln^n(x)}{\sqrt{1-x}}\,dx\end{align}$
Consider,
$\begin{align}B_n(s)&=\frac{1}{2^{n+1}}\int_0^1 x^{-\frac{1}{2}+s}(1-x)^{-\frac{1}{2}}\,dx\\
&=\frac{1}{2^{n+1}}\text{B}\left(\frac{1}{2}+s,\frac{1}{2}\right)\\
&=\frac{1}{2^{n+1}}\frac{\Gamma\left(\frac{1}{2}+s\right)\Gamma\left(\frac{1}{2}\right)}{\Gamma(1+s)}\\
&=\frac{\sqrt{\pi}}{2^{n+1}}\frac{\Gamma\left(\frac{1}{2}+s\right)}{\Gamma(1+s)}
\end{align}$
$\begin{align}\boxed{J_n=\frac{\partial^n B_n(s)}{\partial s^n}\big|_{s=0}}\end{align}$
NB:
$\text{B}(,)$ denotes the Beta Euler function.
Example:
Let $n=1$.
$\begin{align}J_1=\frac{\sqrt{\pi}}{4}\frac{\Gamma^\prime\left(\frac{1}{2}\right)\Gamma(1)-\Gamma\left(\frac{1}{2}\right)\Gamma^\prime(1)}{\Gamma^2(1)}\end{align}$
But,
$\begin{align}\Gamma^\prime\left(\frac{1}{2}\right)=-(\gamma+2\ln 2)\sqrt{\pi}\end{align}$
$\begin{align}\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}\end{align}$
$\begin{align}\Gamma\left(1\right)=1\end{align}$
$\begin{align}\Gamma^\prime\left(1\right)=-\gamma\end{align}$
Therefore,
$\begin{align}J_1&=\frac{\sqrt{\pi}}{4}\times \frac{-(\gamma+2\ln 2)\sqrt{\pi}+\gamma\sqrt{\pi}}{1}\\
&=\boxed{-\frac{1}{2}\pi\ln 2}
\end{align}$
Addendum:
To compute $J_n$ one needs to be able to evaluate $\Gamma^{(k)}(1),\Gamma^{(k)}\left(\frac{1}{2}\right)$ for $0\leq k\leq n$.
Fact 1):
$\begin{align} \Gamma^{(m+1)}(z)=\sum_{k=0}^{m} \binom{m}{k}\Gamma^{(k)}(z)\psi^{(m-k)}(z)\end{align}$
(following from the definition of Digamma function and the use of Leibniz-Newton formula)
Fact 2):
For $0<|z-1|<1$,
$\begin{align}\psi(z)=-\gamma+\sum_{k=1}^{\infty}(-1)^{k+1}\zeta(k+1)(z-1)^k\end{align}$
Therefore,
For $k\geq 1$,
$\displaystyle \psi^{(k)}(1)=(-1)^{k+1}k!\zeta(k+1)$
Fact 3):
For $0<\Re(z)<1$,
$\displaystyle \psi(z)+\psi(1-z)=\pi\cot(\pi x)$
(allowing to compute $\displaystyle \psi^{(k)}\left(\frac{1}{2}\right)$)
A: It's 
$\displaystyle\int\limits_0^{\pi/2} (\sin x)^z dx = {\binom{\frac{z}{2}}{\frac{1}{2}}}^{-1} = 2^{-z} \sin\frac{\pi z}{2} \sum\limits_{k=0}^\infty {\binom z k} \frac{(-1)^k}{z-2k}$ 
with $\enspace\displaystyle (\sin x)^z = \sum\limits_{n=0}^\infty z^n \frac{(\ln \sin x)^n}{n!}\enspace$ , $\enspace \displaystyle 2^{-z} = \sum\limits_{n=0}^\infty z^n \frac{(-\ln 2)^n}{n!}\enspace$ ,
$\displaystyle \frac{1}{z}\sin\frac{\pi z}{2} = \sum\limits_{\,\,n=0\\n\,\text{even}}^\infty z^n \frac{(-1)^{n/2}(\frac{\pi}{2})^{n+1}}{(n+1)!}$
and $\enspace\enspace\displaystyle \sum\limits_{n=0}^\infty z^n a_n := z\sum\limits_{k=0}^\infty {\binom z k} \frac{(-1)^k}{z-2k}\enspace$ , $\enspace$ where $\,a_n\,$ is calculated below.
It follows by coefficient comparison:   

$$ \frac{d^n}{dz^n}{\binom{\frac{z}{2}}{\frac{1}{2}}}^{-1}|_{z=0}=   \int\limits_0^{\pi/2} (\ln \sin x)^n dx = n! \sum\limits_{k=0}^n a_{n-k} \sum\limits_{\,\,j=0 \\ j\,\text{even}}^k \frac{(-\ln 2)^{k-j}}{(k-j)!} \frac{(-1)^{j/2}(\frac{\pi}{2})^{j+1}}{(j+1)!}$$


$a_0=1$ , $\,\,a_1=0$ 
Using the definitions 
$\displaystyle \sum\limits_{k=0}^n \left[ \begin{array}{c} n \\ k \end{array} \right] x^k := x(x+1)…(x+n-1) \enspace\enspace$  (see Stirling numbers of the first kind) $\enspace$ and
$\displaystyle \zeta_n(m):=\sum\limits_{k=1}^\infty \frac{1}{k^m}\left(\frac{n!}{(k-1)!}\left[\begin{array}{c} k \\ n+1 \end{array} \right]\right)\enspace $ 
(as in Evaluate $\int_{0}^{\pi }\theta ^{3}\log^{3}\left ( 2\sin\frac{\theta }{2} \right )\mathrm{d}\theta $)
we get for $\,k\geq 2\,$ the coefficients $\enspace\displaystyle a_k=\sum\limits_{j=1}^{k-1}\frac{(-1)^{j-1}}{2^{k-j}}\frac{\zeta_{j-1}(k-j+1)}{(j-1)!}\enspace$  .
Note: 
$\displaystyle \frac{\zeta_{v}(n-v+2)}{v!}=\frac{\zeta_{n-v}(v+2)}{(n-v)!}\enspace\enspace$ , $\enspace\enspace$ special case $n:=v\,$: $\enspace\displaystyle \frac{\zeta_{v}(2)}{v!}=\zeta(v+2)$
$\displaystyle \frac{\zeta_v(3)}{v!}= \zeta_1(v+2)=(1+\frac{v}{2})\zeta(v+3) - \frac{1}{2} \sum\limits_{j=1}^v \zeta(j+1)\zeta(v+2-j)$
...

In here, part Expansion by harmonic numbers, is $\enspace\displaystyle w(n,m):=\frac{m!}{(n-1)!}\left[ \begin{array}{c} n \\ {m+1} \end{array} \right]\enspace$ with 
the recursion $\enspace\displaystyle w(n,m)= \delta_{m,0}  + \sum\limits_{j=0}^{m-1}(-1)^j \frac{(m-2+j)!}{(m-2)!}H_{n-1}^{(j+1)} w(n,m-1-j) \enspace$ and 
the $\,m$-order harmonic number $\,H_n^{(m)}\,$ so that we can also write $\enspace\displaystyle \zeta_n(m):=\sum\limits_{k=1}^\infty \frac{w(k,n)}{k^m}\,$ .
